Proving Convergence of a Series

[hy23]

Homework Statement

(\sqrt{(n+1)} - \sqrt{n} ) / \sqrt{n}
I'm trying to show this series converges.

Homework Equations

divergence test: as n approaches infinity, if the sequence does not approach 0 then the series diverges
ratio test: as n approaches infinity, if the ratio between subsequent terms is less than 1 then the series converges, if equal to 1 the test is inconclusive, if greater than 1 then the series diverges
comparison test: if this series is smaller than another series that converges, then this series also converges; if this series is greater than another series that diverges, then this series also diverges

The Attempt at a Solution

Applying the divergence test, the sequence definitely approaches 0, but this cannot be used to prove convergence
I tried using ratio test (which has worked so well for nearly every question) and ended up with a mess of square roots.
I'm thinking of using comparison test, but what series can I compare it to?

 

[hunt_mat]

Note that:
<br /> 1=n+1-n=(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})<br />
and hence
<br /> \sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<br />

 

[hy23]

Sorry but I don't see where you're going with this, can you explain?

 

[╔(σ_σ)╝]

He wants you to rationalise the numerator.

 

[hy23]

I know what he did there but I don't know how that step leads to proving convergence

 

[hunt_mat]

Ideally I would like you to compare your series to the harmonic series which is known to diverge

 

[hy23]

yes hunt, but this series is smaller than the harmonic series, thus you can't compare it to the harmonic to prove divergence, maybe it's more plausible to compare it to 1/2n, and use the limit comparison test instead?

 

[╔(σ_σ)╝]

yes hunt, but this series is smaller than the harmonic series, thus you can't compare it to the harmonic to prove divergence, maybe it's more plausible to compare it to 1/2n, and use the limit comparison test instead?

2 \sqrt{n+1} &gt; \sqrt{n} + \sqrt{n+1}

\frac{1}{ 2 \sqrt{n+1}} &lt; \frac{1}{ \sqrt{n} + \sqrt{n+1}}

\frac{1}{2(n+1)} &lt; \frac{1}{ 2 \sqrt{n+1}}

The series diverges.